Question

An ideal capacitor of capacitance $0.2\mu F$ is charged to a potential difference of $10V$. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance $0.5mH$. The current at a time when the potential difference across the capacitor is $5V$, is:

• A. $0.17A$
• B. $0.15A$
• C. $0.34A$
• D. $0.25A$

Answer 1

The correct option is A $0.17A$

Using Energy Conservation,

Initial energy stored in capacitor = Final energy stored in capacitor + Energy stored in inductor

$\frac{1}{2}C{V}_i^2=\frac{1}{2}C{V}_f^2+\frac{1}{2}L{i}^2$

$C(V_i^2-V_f^2)=L{i}^2$

$i=\sqrt{\frac{C(V_i^2-V_f^2)}{L}}$

Here, $V_i=10V$

$V_f=5V$

$C=0.2\mu F$

$L=0.5mH$

On putting these values, we will get $i=0.17A$